Two particles $P$ and $Q$ perform $SHM$ with the same amplitude $a$ and frequency $v$ along the same straight line. The maximum distance between the two particles is $a \sqrt{2}$. The initial phase difference between the particles is:

Consider two SHMs along the same straight line $x_1=A_1 \sin \left(\omega t+\phi_1\right)$ and $x_2=A_2 \sin \left(\omega t+\phi_2\right)$,where $A_1$ and $A_2$ are their amplitudes and $\phi_1$ and $\phi_2$ are their initial phase angles. If $R$ is the resultant amplitude,match the conditions in Column-$I$ with the resultant amplitudes in Column-$II$:

Column-$I$	Column-$II$
$A$. $A_1=A_2=A, \delta=0$	$I$. $A_1+A_2$
$B$. $A_1 \neq A_2, \delta=0$	$II$. $0$
$C$. $A_1=A_2=A, \delta=90^{\circ}$	$III$. $2A$
$D$. $A_1=A_2=A, \delta=180^{\circ}$	$IV$. $A\sqrt{2}$

Two simple harmonic motions are given by $x_{1} = a \sin \omega t + a \cos \omega t$ and $x_{2} = a \sin \omega t + \frac{a}{\sqrt{3}} \cos \omega t$. The ratio of the amplitudes of the first and second motion and the phase difference between them are respectively:

Two particles $p$ and $q$ perform $SHM$ with the same amplitude $a$ and the same frequency $f$ along a straight line. The maximum distance between the two particles is $a\sqrt{2}$. The initial phase difference between the particles is:

Two particles $P$ and $Q$ describe $SHM$ of same amplitude $a$ and same frequency $f$ along the same straight line. The maximum distance between the two particles is $a \sqrt{2}$. The phase difference between the particles is

The displacement of a particle in a periodic motion is given by $y = 4 \cos^{2}\left(\frac{t}{2}\right) \sin(1000 t)$. This displacement may be considered as the result of the superposition of $n$ independent harmonic oscillations. Here $n$ is